Limits And Derivatives Class 11 Notes Mathematics Chapter 13 - CBSE

Chapter : 13

What Are Limits And Derivatives ?

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    Limits

    The expected value of function as dictated by the points to the left of a point defines left hand limit of a function at that point. Similarly, right hand limit. Limit of a function at a point is common value of left and right hand limits, if they coincide.

    $$\lim_{x\to a}\space f(x)\space\text{exists}\Leftrightarrow\\\lim_{x\to a}\space f(x) = \lim_{x\to a}f(x)$$

    For a function f(x) and a real number a

    $$\lim_{x\to a}\space f(a)\space\text{and f(a)}\\\text{may not be same.}$$

    Facts

    $$\centerdot\lim_{x\to 0}\space f(x)\space\text{exists but f(a)}\\\text{[the vaule of f(x) at a]}\\\text{may not exists.}\\\centerdot\space\text{The value of f(a) exists}\\ \text{but\space}\lim_{x\to a} f(x)\space \text{does not exists.}\\\centerdot \lim_{x\to a}\space \text{f(x) and f(a)}\\\text{both exist but are unequal.}\\\centerdot \lim_{x\to a}\space\text{f(x) and f(a)}\\\text{both exist but are equal.}$$

    Alegbra Of Limits

    $$\text{Let}\space \lim_{x\to a}\space f(x) \text{and} = \text{I and}\\\lim_{x\to a}g(x) = m.$$

    If l and m both exists then

    $$\centerdot\space\lim_{x\to a}\space \text{kf(X)} = k\lim_{x\to a}f(x)\\\centerdot\lim_{x\to a}(f\pm g)(x)=\\\lim_{x\to a} f(x)\pm \lim_{x\to a} g(x) =l\pm m\\\centerdot \lim_{x\to a}\space(fg)(x) =\\\lim_{x\to a} f(X).\lim_{x\to a} g(x) = lm\\\centerdot\space\lim_{x\to a}\frac{f}{a}(x) =\frac{\lim_{x\to a} f(x)}{\lim_{x\to a} g(x)}=\frac{1}{m},\\\text{Here g(x)} ≠ 0\\\centerdot\space\lim_{x\to a}\lbrace f(x)\rbrace^{g(x)} = \text{I}^{\text{m}}$$

    Standard Limits

    $$\centerdot\space\lim_{x\to a}\frac{x^{n}- a^{n}}{x-a} = na^{n-1}\\\centerdot\space\lim_{x\to 0}\frac{\text{sin x}}{x}= 1\\\centerdot\lim_{x\to0}\frac{\text{tan x}}{x}=1\\\centerdot\lim_{x\to a}\frac{\text{sin(x-a)}}{x-a} = 1\\\centerdot \lim_{x\to a}\frac{\text{tan}(x-a)}{x-a} = 1\\\lim_{x\to 0}\frac{\text{log}(1 +x)}{x} = 1\\\centerdot\lim_{x\to0}\frac{a^{x}-1}{x} =\text{log}_{e}a,\\ a\neq0, a\gt 1$$

    $$\centerdot\space\lim_{x\to 0}\frac{e^{x}-1}{x} = 1\\\centerdot\space\lim_{x\to0}\frac{\text{1 - cos x}}{x} =1$$

    Derivative At A Point

    Let f(x) be a real valued function defined on an open interval (a, b) and let c Î (a, b). Then, f(x) is said to be differentiable or derivative at x = c if

    $$\lim_{x\to c}\frac{f(x) - f(c)}{x-c}\\\text{exists finitely.}$$

    This limit is called the derivative or differentiation of f(x) at x = c and denoted by f'(c) or Df(c) or

    $$\begin{Bmatrix}\frac{d}{dx}f(x)\end{Bmatrix}_{x = c}$$

    If f(x) is a differentiable function, then

    $$\lim_{h\to 0}\frac{f(x+h)f(x)}{h}\space\\\text{is called the differentiation of f(x)}$$

    or differentiation of f(x) with respect to x.

    Standard Derivatives

    $$\centerdot\space \frac{d}{dx}(x^{n}) = nx^{n-1}\\\centerdot\frac{d}{dx}(a^{x}) = d^{x}\space\text{log}_ea,\\a \gt0, a\neq 1\\\centerdot\frac{d}{dx} e^{x} = e^{x}\\\centerdot \frac{d}{dx}(\text{log}_{e}x)=\frac{1}{x}\\\centerdot\frac{d}{dx}(\text{sin x}) = cos\space x\\\centerdot \frac{d}{dx}(\text{cos x}) = -\text{sin x}\\\centerdot\frac{d}{dx}(\text{tan x}) =\text{sec}^{2}x$$

    $$\centerdot\frac{d}{dx}(\text{cot x}) =-\text{cosec}^{2} x\\\centerdot\frac{d}{dx}(sec \space x) =\text{sec x tan x}\\\centerdot\frac{d}{dx}(\text{cosec x}) =-\text{cosec x cot x}$$

    Fundamental Rules For Differentiation

    • Differentiation of a constant function is zero. i.e.

    $$\frac{d}{dx}(c) = 0.$$

    • Differentiation of a constant and a function is a equal to constant times the differentiation of a function.
    • If f(x) and g(x) and differentiable functions, then

    $$\textbf{(a)}\frac{d}{dx}\lbrack f(x)\pm g(x)\rbrack=\\\frac{d}{dx}f(x)\pm\frac{d}{dx}g(x)\\\textbf{(b)}\frac{d}{dx}|\text{f(x) × g(x)}|=\\ f(x)\frac{d}{dx} g(x) + g(x)\frac{d}{dx}f(x)\\\textbf{(c)}\frac{d}{dx}\begin{Bmatrix}\frac{f(x)}{g(x)}\end{Bmatrix}=\\\frac{g(x)\frac{d}{dx}f(x) - f(x)\frac{d}{dx}g(x)}{[g(x)]^{2}}\\\lbrace\because\space g(x)\neq 0\rbrace$$

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